Problem: Simplify the following expression and state the condition under which the simplification is valid: $z = \dfrac{r^2 - 7r}{r^2 - 5r - 14}$
Answer: First factor the expressions in the numerator and denominator. $ \dfrac{r^2 - 7r}{r^2 - 5r - 14} = \dfrac{(r)(r - 7)}{(r + 2)(r - 7)} $ Notice that the term $(r - 7)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(r - 7)$ gives: $z = \dfrac{r}{r + 2}$ Since we divided by $(r - 7)$, $r \neq 7$. $z = \dfrac{r}{r + 2}; \space r \neq 7$